How do you find an equation of hyperbola with given foci and asymptote: Foci: (0,0) and (0,4) Asymptote: y= +/-[(1/2)]x + 2?

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Answer 1 · 2017-08-09T15:03:18

The equation of the hyperbola is $(y-2)^2-(x^2/4)=1$

The foci are $F=(0,4)$ and $F'=(0,0)$

The center is $C=(0,2)$

The equations of the asymptotes are

$y=1/2x+2$ and $y=-1/2x+2$

Therefore,

$y-2=+-1/2x$

Squaring both sides

$(y-2)^2-(x^2/4)=0$

Therefore,

The equation of the hyperbola is

$(y-2)^2-(x^2/4)=1$

Verification

The general equation of the hyperbola is

$(y-h)^2/a^2-(x-k)^2/b^2=1$

The foci are $F=(k,h+c)=(0,2+2)=(0,4)$ and

$F'=(k,h-c)=(0,2-2)=(0,0)$

graph{((y-2)^2-(x^2)/4-1)(y-2-1/2x)(y-2+1/2x)=0 [-6.76, 7.28, -1.425, 5.6]}